Radiation from an Accelerated Charge

Abhishek DasguptaChallenger Mishra

8th December 2009

Abstract

In this we calculate the radiated power from an accelerated chargewhich has just come to rest using geometry of electric field lines. Theresult applies to a wide variety of systems from radio antennas to atomsand nuclei.

1 The problem

We consider a charged particle which was moving with a velocity v0 ! c fort < 0. At t = 0 it starts decelerating uniformly and comes to a stop in time! . The distance covered in this time is 1

2v0! . We’ll find out how the electricfield looks like at a given time t = T " ! . As electromagnetic waves travelat the speed of light c, observers farther away from the origin than R = cTwould not be aware of the particle’s deceleration. Throughout this region(R > cT ) which we call region I, the electric field is that of a charge movingwith constant velocity v0. The field lines would appear to emanate from theposition x = v0T which is the distance the particle would have travelled hadit not decelerated. However for any observer who is within the spherical shellof radius c(T # !) (in region II), the field is that of a charge at rest close tothe origin at x = 1

2v0! (fig 1)

2 How to join the field lines?

It is a relatively simple matter to connect up the inner and outer field lines.There is only one way to do this that is consistent with Gauss’ law. We start

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Figure 1: A charged particle brought to rest by constant deceleration startingat t=0

with some point A in region II on the radial field line making angle "0 withthe x-axis till we reach the R = c(T # !), follow the field line till we emergeout into region I and continue on a line making an angle #0 with the x-axis(fig 1). If we rotate EABCDF about the x axis we would generate a surface ofrevolution which encloses no charge. By applying Gauss’ law to this surfacewe would derive a relationship between "0 and #0 . The contribution to theintegral

´

$E. $dS would come from the caps formed by revolving AE and DFabout x axis (fig 2). The flux through the inner cap is

ˆ !0

0

q

r22%r2 sin "d" = 2%q

ˆ !0

0sin "d" (1)

and through the outer cap

ˆ "0

0

q

r2

1# &2

(1# &2 sin2 #)3/22%r2 sin#d# = 2%q

ˆ "0

0

1# &2

(1# &2 sin2 #)3/2sin#d#

(2)

where we’ve used the form of the electric field for a uniformly moving chargedparticle viz.

E =q

r2

1# &2

(1# &2 sin2 ")3/2(3)

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Figure 2: Only the inner and the outer cap contribute to the surface integralof $E

where r and " are measured relative to the position of the charge. The fluxinto the inner cap is the flux out through the outer cap, and hence

ˆ !0

0sin "d" =

ˆ "0

0

1# &2

(1# &2 sin2 #)3/2sin#d# (4)

which simplifies to

cos "0 =cos #0!

1# &2 sin2 #0

(5)

and equivalently to

tan#0 = ' tan "0 (6)

This reduces to "0 = #0 under the condition ' $ 1 i.e., for non-relativisticspeeds.

3 Using the geometry of the electric field lines

The only problem lies in finding the electric field in the annular region be-tween R = c(T # !) and R = cT . We use Gauss’ law. We have already seen

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Figure 3: Space diagram for t = T " !

that AB and CD must be part of the same field line connected by BC. Alsowe have "0 = #0 = " since v0 ! c. The electric field within the shell whichhas both radial and transverse components. From the geometry of the fieldlines (fig 3) in this annular region

E!

Er=

v0T sin "

c!(7)

Taking a small Gaussian pillbox near B, we see that Er has the same valuewithin the shell thickness as in region II at B. Thus Er = q/R2 = q/c2T 2.Substituting, we get

E! =v0T sin "

c!Er =

qv0 sin "

c3T !(8)

Rewriting using the results v0/! = a and R = cT

E! =qa sin "

c2R(9)

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It is remarkable that E! in contrast to Er is proportional to 1/R instead of1/R2. With increasing time and thus increasing R, the transverse field willthus become much stronger than the radial field. Along with this transverseelectric field, there will also be a magnetic field orthogonal to E which canbe found out by applying Ampere’s law in the integral form.

4 Power radiated from an accelerating charge

Now we shall find the energy stored in the electric field in the whole sphericalshell. The energy density for the transverse field is (we are ignoring the radialcomponent as it can be neglected in comparison to the transverse for largeR)

E2!

8%=

q2a2 sin2 "

8%R2c4(10)

The volume of the shell is 4%R2c! and the average value of sin2 " over thewhole sphere is 2/3. We can see this by taking the polar axis as the xaxis (x2 = R2 cos2 "). Due to symmetry,

"x2

#=

"y2

#=

"z2

#= R2/3.

Thus"cos2 "

#= 1/3, implying

"sin2 "

#=2/3. Then the total energy of the

transverse electric field is

234%R2c!

q2a2

8%R2c4=

q2a2!

3c3(11)

We add an equal amount of energy stored in the magnetic field to get thetotal energy 2q2a2!/3c3. As there is no dependence on R, this energy simplytravels out undiminished with the speed of light c. An observer would observethe radiation as a pulse with the time width ! . Thus the power radiatedduring the accelration process is

Prad =23

q2a2

c3(12)

This is a Lorentz invariant quantity since Prad is energy/time, and energytransforms like time, each being the fourth component of a four-vector. Thisexpression gives the instantaneous rate of radiation by a charged particlemoving with variable acceleration.

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References

[1] Purcell E M 1981 Electricity and Magnetism, Berkeley Physics Coursevol 2 (Asian Students edition) (Singapore: McGraw-Hill)

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